Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$
Then,
(A)$|x_i|=2$ for exactly one value of $i$
(B)$|x_i|=2$ for exactly two values of $i$
(C)$|x_i|=2$ for all values of $i$.
(D)$|x_i|=2$ for no value of $i$.
My attempt: I noticed that the polynomial forms a geometric
progression with common ratio $2/x$.
Then I summed the terms and found this relation $x^7=128$ which gives us
$x=2$ but $x=2$ doesn't satisfy the original polynomial.
I looked at the graph too and found that the polynomial has no real roots .
So now , how to find the modulus of the roots ?
Best Answer
$$f(x)=\sum_{r=0}^6(x^r2^{6-r})=2^6\dfrac{\left(\dfrac x2\right)^7-1}{\dfrac x2-1}=\dfrac{x^7-128}{x-2}$$
Clearly, $f(2)\ne0$
$\implies x^7=128$ with $x\ne2$
$\iff\left(\dfrac x2\right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$